Question:

The wire splits into two ways, which are bent circularly with radius of \(a\). The upper half wire has resistance \(2R\) and the lower part has \(R\). Find the magnetic fields at the center of the circle in terms of the total current \(I\).

Answer:

The magnetic field created by current \(I_1\) is denoted as \(B_1\). From Biot-Savart law,

\[\vec{B_1} = \oint \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times \vec{r}}{r^2}  \]

\[    = \frac{\mu_0}{4\pi}\frac{I_1}{a^2}\int_{0\rightarrow \pi a}dl (-\vec{z})  \]

\[    = -\frac{\mu_0}{4\pi}\frac{I_1}{a^2}\pi a\vec{z} \]

\[    = -\frac{\mu_0 I_1}{4a}\vec{z}  \]

Likewise, we can obtain the magnetic field created by the lower part.

\[  \vec{B_2} = \frac{\mu_0 I_2}{4a}\vec{z}  \]

This is a parallel connection, and we can use the reciprocal relationship between currents and resistances:

\[  \frac{I_1}{I_2}=\frac{R}{2R} \quad \rightarrow \quad I_1=\frac{1}{2}I_2  \]

The current is conserved.

\[  I = I_1 + I_2 \quad \rightarrow \quad I_2 = I – I_1  \]

Therefore,

\[  I_1=\frac{1}{2}(I – I_1) \quad \rightarrow \quad I_1 = \frac{I}{3}  \]

Hence, the other current will be

\[  I_2 = \frac{2I}{3}  \]

The magnetic field is expressed as

\[  \vec{B}=\vec{B_1}+\vec{B_2}=\frac{\mu_0}{4a}(I_2-I_2)\vec{z}=\frac{\mu_0}{4a}\left(\frac{2I}{3}-\frac{I}{3}\right)\vec{z}=\frac{\mu_0 I}{12a}\vec{z}  \]

The magnetic field is directed toward us from the monitor.